Counting Rules

1 Stat 101/math 107

Department of Statistics
Forman Christian College (A Chartered University), Lahore
2 Counting Rules

Addition Rule
 Many times a person must
know the number of all
possible outcomes for a
sequence of events. Multiplication Rule

 To determine this number,

four rules can be used
Permutation Rule

Combination Rule
3 Addition Rule
 If a task can be done in n1 ways and a second take in n2
ways, and if these two tasks cannot be done at the same
time, then there are n1 and n2 ways to do either task.

 Example 1: If you are going to have icecream or pie for

dessert, there are two choices (vanilla or butter pecan)
if you have icecream, and three choices (apple, cherry, or
blueberry) if your have pie.

 Then there are 2+3= 5 choices for dessert

.
If you are going to choose from one catergory or another category,
then you add the number of choices in each category.
4 Example 2:

 Suppose,we have to write an article and

we have two list of topcis
 First list contains 6 topics
 Second Contains 8 Topics

Total number of possible options of

topics are 6+8= 14
We can choose any one option of them
5 Example 3:
 Suppose that either a member if the mathemtics fcaulty or a
student who is a mathemtics major is chosen as a
representative to a Univeristy Committee. How many different
choices are there for this representative if there are 37
members of the mathemtics faculty and 83 mathemtics
majors?

37 83 120
 So there are 120 possible ways to pick this representative
6 The Multiplication Rule

 In a sequence of n operations in which the first one has k1 possibilities and

the second event has k2 possibilities of the sequence will be
𝑘1 × 𝑘2 × 𝑘3 × ⋯ × 𝑘𝑛
 If a woman has three skirts and four sweaters, how many outfits are possible.
 Answer:
Skirts has 3 possibilities = 𝑘1
Sweaters has 4 possibilities= 𝑘2

𝑘1 × 𝑘2 = 3× 4 = 12
There are 12 possiblbe outfits a woman can wear.

 When one must make a choice from one category and a choice from another
category, one multiplices the number of choices in the two categories.
  Example 4: A paint manufacturer wishes to manufacture several
7 different paints. The categories include

Colour Red, blue, white, black, green, brown, yellow

Type Latex, Oil
Texture Flate , semigloss, high gloss
Use Outdoor, indoor

 How many different kinds of paint can be made if you can select
one colour, one type, one texture, and one use?
  Solution
8
You can choose one colour and one type and one
texture and one use.
Since there are 7 colour choices, 2 type choices, 3
texture choices, and 2 use choices, the total number of
possible different paints is

Color Type Texture Use

7 × 2 × 3 × 2 =84
9 Example 5:
Joe has 7 shirts, 4 pairs of pants and 2 pairs of shoes. He needs to
make an outfit containing one of each item. How many different
outfits are possible?
7- shirts
4-pairs of pants
2 pairs of shoes
Total possibilities =

7 4 2 56
Example 3 video link:
https://www.youtube.com/watch?v=549eLWIu0Xk&t=610s
 Practice Questions
10

There are four blood types, A, B, AB, and O.

1
Blood can also be Rh+ and Rh-. Finally, a
blood donor can be classified as either male
or female. How many different ways can a
donor have his or her blood labelled?

A simple survey consists of three multiple

choice questions. The first question has 3

2 possible answers, the second has 4 possible

answers and the third has 3 possible answers.
What is the total number of different ways in
which this survey could be completed?

There are 21 novels and 18 volumes of

3
poetry on a reading list for a college English
course. How many different ways can a
student select one novel and one volume
of poetry to read during the quarter?
 Multiplication Rule with tree diagram
11  Example 6: there are 3 egg choices (scrambled, fried, omelete) and two meats
choices (sausage and chicken patty). How many different breakfast do you have?
3 outcomes 2 Outcomes 6 outcomes (total)

Sausage Scrambled-Sausage
Scrambled
Chicken patty Scrambled- chicken
patty

Sausage Fried-Sausage
fried
Chicken patty Fried-chicken patty

Sausage Omelet-Sausage
Omelete
Chicken patty
Omelet-chicken patty
 Example 7: A room has three doors marked A, B,C. In how many
12 ways can a person enter by one door and leave by a different door?

Solution 3 outcomes 2 outcomes = 6 outcomes

Enter A- Enter
A
Exit A- Exit

Enter B- Enter
B
Exit B- Exit

Enter C- Enter
C
Exit C- Exit
13 Practice Questions

1
Using a tree diagram, determine the
number of outcomes when rolling a six-
sided die and flipping a coin. Check
your work using the counting principle.

2
Using a tree diagram, determine the
number of outcomes when rolling a six-
sided die and flipping two coins. Check
your work using the counting principle

Shannon is going to purchase a new car.

3
The car has 5 exterior colour choices, 2
interior colour packages and 2 engine
options. How many different cars are
possible?
14 Example 8

 If two dates are selected at random from the 365 days of the
year and are listed in the order they were selected.

a) Use the counting principle to determine the number of

possible outcomes if the dates are selected with
replacement.
b) Use the counting principle to determine the number of
possible outcomes if the dates are selected without
replacement.
15 Solution

(a) with replacement (b) without replacement

(with repitition) (without repitition)

365
365
133225
132860
days days
365
364
 Factorial Notation
16

 These rules use factorial Notation. The factorial notation uses the
exclaimation point .
 For any counting n
n! = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ × 1

 5! = 5 × 4 × 3 × 2 × 1
 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
 Factorial Notation
17

 To use the formulas in the permutation and combination rules,

a special definition of 0! Is needed.
0! = 1

 Example video:
https://www.youtube.com/watch?v=regctIrmOGA
 Permutations
18

 A permutation is an arrangment of n objects in a specific order.

 Example 9: Suppose a business owner has a choice of 5 locations in

which to establish her business. She decides to rank each location
according to certain criteria, such as price of the store and parking
facilities. How many different ways can she rank the 5 locations?

 Solution : there are

5! = 5 × 4 × 3 × 2 × 1 = 120
Different possible rankings. The reason is that she has 5 choices for the first
location, 4 choices for the second location, 3 choices for the third location, etc.
 19

Example 10:
Suppose the business owner in
example 5 wishes to rank only the
top 3 of the 5 locations. How many
different ways can she rank them?
 Using the multiplication counting
rule, she can select any one of the 5
for first choice, then any one of the
remaining 4 locations for her second,
finally, any one of the remaining
locations for her third choice, as
shown.
5 4 3 60
20 Permutation Rule-1 (without repitition)

 Rule -1:The arrangement of n objects in a specific order using

r objects at a time is called a permutation of n objects taking
r objects ar a time. It is written as nPr and the formula is

nP
𝑛!
r =
𝑛−𝑟 !
 The notation nPr is used for permutations.
6P
6! 6! 6×5×4×3×2×1
 4 means = = = 360
6−4 ! 2! 2×1
 Example 11:
21 A television news director wishes to use 3 news stories on an
evening show. One story will be the lead story, one will be the
second story, and the last will be a closing story. If the director
has a total of 8 stories to choose from, how many possible ways
can the program be set up?
 Solution :
𝑛!
 nPr =
𝑛−𝑟 !

8P 8! 8! 8×7×6×5!
 3 = = = = 336
8−3 ! 5! 5!

Hence, there would be 336 ways

to set up the program
 Example 12:
22 A school musical director can select 2 musical plays to present next
year. One will be presented in the fall, and one will be presented in the
spring. If she has 9 to pick from, how many different possibilities are
there?

Solution:
 Order is important since one
play can be presented in the
fall and the other play in the
spring
nP =
𝑛!
r
𝑛−𝑟 !
9P =
9! 9! 9 × 8 × 7!
2 = =
9 − 2 ! 7! 7!
= 72
 There are 72 different
possibilities
23 Practice Questions
How many different 4-color code stripes

1 can be made on a sports car if each code

consists of the colours green, red, blue, and
white? All colours are used only once.

How many different 4-letter permutations can

2 be formed from the letters in the word
decagon

How many different ways can 4 tickets be

3 selected from 50 tickets if each ticket wins a
different prize?

How many different ID cards can be made if

4 there are 6 digits on a card and no digit can
be used more than once?
 Permutation with Tree Diagram
24
Example13: Find the permutations for the letters of the following three words
using formula and tree diagram
(a)HI (b) JET (c)MOM

2 letters 1 letter 2 words

Solution (a)
nP
𝑛!
=

Tree diagram
r
𝑛−𝑟 !
formula

I HI
2!
= 2−2 !
= 2!= 2 words

H IH
25
Solution
(b) JET
3 letters 2 letters 1 letter 6 words

JET
E T
 nPr =
𝑛! J
JTE
formula

𝑛−𝑟 !
T E

Tree diagram
3!
= = 3!=6
3−3 !
EJT
J T
E
T J
ETJ

J E TJE
T
E J
TEJ
 M MOM
26 (c) MOM O
M
O MMO
NOTE: M
in this word “MOM”
M is repeating twice. Which
results into 3 unique mixture

SAME WORDS
M OMM
of words. M

Tree diagram
So we cannot apply the O
M OMM
permutation Rule-1 (without M
repetition) here

M MOM
O
M
Example video link:
O MMO
https://www.youtube.com/watch? M
v=v-4WTBk8a88
The same color in last column shos same words
27 Permutations Rule-2
 In the previous examples, all items involving permutations
were different, but when some of the items are identical, a
second permutation rule can be used.

 Rule-2: The number of permulations of n objects are r1

objects are identical, r2 objects are identical, ……, rp objects
are identical etc., is
𝑛!
𝑟1 ! 𝑟2 ! … 𝑟𝑝 !
where 𝑟1 + 𝑟2 +…+ 𝑟𝑝 = n
28 Example 14:

 Mrs. Cottrell has 9 old

yearbooks on her shelf, 4 are
from 2015, 2 are from 2014, 1
is from 2013 and 2 are from
2012. How many different
ways can she order the
yearbooks on her shelf?
𝑛! 9!
=
𝑟1 ! 𝑟2 ! … 𝑟𝑝 ! 4! 2! 1! 2!
= 3780 𝑤𝑎𝑦𝑠
 Example 15:
29

 How many different ways can we rearrange the

letters of MISSISSIPPI?
 We have 11 letters in total, of which
1 is ‘M’
4 are ‘I’
4 are ‘S’
‘2’ are ‘P’
In this situation, the total number of different
rearrangements
𝒏! 𝟏𝟏!
= = 𝟑𝟒𝟔𝟓𝟎 𝒘𝒂𝒚𝒔
𝒓𝟏 ! 𝒓𝟐 ! … 𝒓𝒑 ! 𝟏! 𝟒! 𝟒! 𝟐!
 Practice Questions:
30 Find the number of distinguishable permutations of the letters in
the word. (Also use tree-diagram)

 Permutation without Repetition  Permutation with Repitition

1. PENCIL 1. ALABAMA
2. TEXAS 2. CONNECTCUT
3. GOLDFISH 3. MISSOURI
4. FLORIDA 4. LETTER
 Combinations
31

 A selection of distinct objects without regard to order is called a

combination
• This is different from a permutation because in a combination,
order does not matter.
 The difference between a permutation and a combination can be
seen in a set of four letters { A,B,C,D} where two are chosen.

Order matters Order does not matters

Permutations Combinations
(AB),(BA), (AC), (CA), (AD),(DA), (AB),(AC), (AD), (BC), (BD), (CD)
(BC), (CB), (BD), (DB), (CD), (DC)

(AB)and (BA) are different permutations. (AB) and (BA) are the same combination
 Combination Rule
32
 Combinations are used when the order or
arrangement is not important, as in the
selecting process.

 Suppose a committee of 5 students is to be

selected from 25 students. The five selected
students represent a combination, since it
does not matter who is selected first, second,
etc.
 The number of combinations of r objects
selected from n objects is denoted by nCr and
is given by the formula
nC =
𝑛!
r
(𝑛 − 𝑟)! 𝑟!
  Solution
33

nC =
𝒏!
r
Example 16: (𝒏 − 𝒓)! 𝒓!
A newspaper editor has received 8
books to review. He decides that he
nC =
𝟖!
can use 3 reviews in his newspaper. r
How many different ways can these 3 (𝟖 − 𝟑)! 𝟑!
reviews be selected? nC =
𝟖!
r = 𝟓𝟔
𝟓! 𝟑!

There are 56 possibilities

 Example 17:
34 An advertising executive must select 3 different photographs for an
advertising flier. If she has 10 different photographs that can be used,
how many ways can she select 3 of them?

 Solution
nC
𝑛!
r=
(𝑛 − 𝑟)! 𝑟!

nC =
10!
r
(10 − 3)! 3!
nC =
8!
r = 120
7! 3!

There are 120 combinations to slect 3

photographs

Video Example: https://www.youtube.com/watch?v=lFlJkINbUbA

35 Practice Questions

How many ways are there to

1 select 3 bracelets from a box of

10 bracelets, disregarding the
order of selection?

How many ways can a

2 committee of 4 people be
selected from a group of 10
people?

If a person can select 3 presents

3 from 10 presents under a Christmas

tree, how many different
combinations are there?
 Combinations with
36
Multiplication Rule
 Example 18:In a club there are 7 women and 5
men. A committee of 3 women and 2 men is to
be chosen. How many different possibilities are
there?
 Here, you must select 3 women from 7 women,
which can be done in 7C3 = 35, ways.
 Next, 2 men must be selected from 5 men,
which can be done in 5C2,=10, ways.
 Finally, by the fundamental counting rule, the
total number of different ways is
7C × 5C =
7! 5!
3 2 × = 35 × 10 = 350
(7 − 3)! 3! (5 − 2)! 2!
 Example19:
37

 Coleen is on Shopping sprre. She buys six

tops, three skorts and 4 pairs of samdals.
How many different outfits consisting of a
top, shorts and samdals can she create
from here new purchases?
nC
𝑛!
r=
(𝑛 − 𝑟)! 𝑟!

6C 3C 4C
6! 3! 4!
1 × 1 × 1 = × ×
(6 − 1)! 1! (3 − 1)! 1! (4 − 1)! 1!
=6×3×4
= 72 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑓𝑖𝑡𝑠
 Practice Questions
38

How many ways can 4 baseball

1
players and 3 basketball players
be selected from 12 baseball
players and 9 basketball
players?

In a train yard there are 4 tank cars,

12 boxcars, and 7 flatcars. How

2 many ways can a train be made up

consisting of 2 tank cars, 5 boxcars,
and3 flatcars? (In this case, order is
not important.)

3
How many ways can a foursome of 2
men and 2 women be selected from
10 men and 12 women in a golf club?
39 Combinations with addition Rule
 Example 20 : A committee of 5 persons is to be formed from 6 men and 4
women. In how many ways can this be done when
(i) at least 2 women are included? (ii ) atmost 2 women are included?
 Solution
(i) When at least 2 women are included .
The committee may consist of
2 women, 3 men : it can be done in 4C2 × 6C3 ways
Or 3 women, 2 men : it can be done in 4C3 × 6C2 ways
Or 4 women, 1 man : it can be done in 4C4 × 6C1 ways
 Therefore, total number of ways of forming the committee
= 4C2 × 6C3 + 4C3 × 6C2 + 4C4 × 6C1
= 6 × 20 + 4 × 15 + 1 × 6
= 120 + 60 + 6 = 186
40

(ii) When atmost 2 women are included .

The committee may consist of
2 women, 3 men : it can be done in 4C2 × 6C3 ways
Or 1 woman, 4 men : it can be done in 4C1 × 6C4 ways
Or 0 women, 5 men : it can be done in 4C0 × 6C5 ways
 Therefore, total number of ways of forming the committee
= 4C2 × 6C3 + 4C1 × 6C4 + 4C0 × 6C5

= 6 × 20 + 4 × 15 + 1 × 6

= 120 + 60 + 6 = 186
41

To fill up 5 vacancies, 25 applications

The Indian Cricket team consists of16 were recieved. There were 7 S.C. and 8
players. It includes 2 wicket keepers O.B.C. candidates among the applicants.
and 5 bowlers. In how many ways can If 2 posts were reserved for S.C. and 1 for
a cricket eleven be selected if we O.B.C. candidates, find the number of
have to select 1 wicket keeper and at ways in which selection could be made?
least 4 bowlers?
Practice
Questions

The Hawaiian alphabet consists of

Out of 5 men and 3 women, a committee 7 consonants and 5 vowels. How
of 3 persons is to be formed. In how many
ways can it be formed selecting (i) many three-letter “words” are
exactly 1 woman. (ii) atleast 1 woman. possible if there are more than two
consonants together?